Equation For Capacitor Leakage Current

Question

I'm finding a bit of a problem when trying to solve this question:

Show that the leakage current for the capacitor is given by \$ I = \dfrac{Q}{KP} \$

where \$Q\$ is the charge on the plates, \$K\$ is the dielectric constant and \$P\$ is the resistivity of the dielectric.

I know the concept behind the leakage current and why it happens, but I don't know how I can prove the equation.

@DaveTweed it's been corrected now, originally the formula said K = Q/(KP) — Andy aka, Commented May 1, 2014 at 14:33

Dave Tweed · Accepted Answer · 2014-05-01 14:31:52Z

0

Current is defined as voltage divided by resistance.

Therefore, you need to find a way to relate charge to voltage, and resistivity to resistance.

Voltage on a capacitor is charge divided by capacitance. Capacitance is related to plate area, spacing and dielectric constant.

Resistance is resistivity multiplied by thickness and divided by area.

Are these hints enough to get you started?

answered May 1, 2014 at 14:31

Dave Tweed

178k17 gold badges242 silver badges418 bronze badges

\$\begingroup\$ Thank you, My equation simplified down to I=dQ/EPL. Where d is the distance between the capacitor plates, Q is the charge on the plates, E is the Permittivity, P is resistivity and L is the thickness. I cant figure out how can i simplify it more to get I=Q/(KP) Is there a relation between K, d, E, and L? \$\endgroup\$
– Mohdak
Commented May 1, 2014 at 14:58
\$\begingroup\$ "Dielectric constant" normally refers to the relative permittivity \$\epsilon_r\$ of the dielectric (relative to vacuum), and is a multiplier used with \$\epsilon_0\$, the permittivity of free space. In this case, I think K refers to the total permittivity of the dielectric, i.e., \$\epsilon_r\epsilon_0\$. This means that your E and their K are the same thing. Also, note that d and L are the same thing, so they cancel out. \$\endgroup\$
– Dave Tweed
Commented May 1, 2014 at 15:15

Add a comment |

OhMyGauss · Accepted Answer · 2020-04-15 07:38:36Z

-1

\$ \vec{j} = \sigma \vec{E} \$

\$ i = \displaystyle \int\vec{j}.\vec{ds}\$

\$ i = \sigma\displaystyle \int\vec{E}.\vec{ds}= \sigma E \displaystyle \int {ds} = \sigma E*S\$

\$ i=\sigma\frac{qS}{k\epsilon_o S} \$

answered Apr 15, 2020 at 7:38

OhMyGauss

99

\$\begingroup\$ correct, but maybe you should explain this in a simpler way, Mohdak has been having trouble with a simple equation, adding surface integrals without any comment may make it even more confusing to him even though its basic physics \$\endgroup\$
– diegogmx
Commented Apr 15, 2020 at 12:39
\$\begingroup\$ @diegogmx sure. The current density \$ \vec{j} \$ is inside the dielectric and since the field inside the dielectric is constant and is alon the infinitesimal surface area vector, \$ \vec{E} \$ comes out of the integral and the surface area is the resultant integral. That's pretty much it about the calculations for the leak current. If you want to calculate the charge as a function of time, then put \$ i = -\frac{dq}{dt} \$ and solve it with the obtained expression of current. \$\endgroup\$
– OhMyGauss
Commented Apr 15, 2020 at 14:47
\$\begingroup\$ you know what's interesting, i didn't downvote your answer, it was someone else i can't remove the downvote, i just posted a friendly comment. Nevertheless i'm flagging your comments and suggesting you to seek professional help, it's not normal nor healthy to react in that way for a stupid stack exchange vote \$\endgroup\$
– diegogmx
Commented Apr 17, 2020 at 14:20
\$\begingroup\$ @diegogmx as if I care about your stupid flag. Do what your stupid mind pleases \$\endgroup\$
– OhMyGauss
Commented Apr 18, 2020 at 8:18

Add a comment |

2 Answers 2